n you list the steps on how to s
Elementary Algebra
17th Edition
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:Lynn Marecek, MaryAnne Anthony-Smith
Chapter2: Solving Linear Equations And Inequalities
Section2.1: Solve Equations Using The Subtraction And Addition Properties Of Equality
Problem 2.17TI: Translate and solve: Ten more than x is equal to 41.
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can you list the steps on how to solve this (how to get to the answer), please.
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### Calculus Problem: Related Rates
**Problem Statement:**
The length of a rectangle is increasing at a rate of \(6 \, \text{cm/s}\) and its width is increasing at a rate of \(3 \, \text{cm/s}\). When the length is \(13 \, \text{cm}\) and the width is \(9 \, \text{cm}\), how fast is the area of the rectangle increasing?
\[ \text{Rate of increase of area} = \boxed{ \text{cm}^2/\text{s} } \]
---
**Explanation:**
This problem involves related rates, a common type of problem in calculus where we determine how one quantity changes as another quantity changes. To solve this problem, you must understand how the dimensions of a geometric shape (in this case, a rectangle) relate to its area and how changes in these dimensions affect the area over time.
1. **Identify Variables:**
- Let \( l \) represent the length of the rectangle.
- Let \( w \) represent the width of the rectangle.
- Let \( A \) represent the area of the rectangle.
2. **Given Rates:**
- The rate of change of length \( \frac{dl}{dt} = 6 \, \text{cm/s} \)
- The rate of change of width \( \frac{dw}{dt} = 3 \, \text{cm/s} \)
3. **Formula for the Area of Rectangle:**
- The area of the rectangle is \( A = l \times w \)
4. **Differentiate with Respect to Time:**
- Use the product rule to differentiate the area with respect to time.
\[
\frac{dA}{dt} = \frac{d}{dt} (l \times w) = l \frac{dw}{dt} + w \frac{dl}{dt}
\]
5. **Substitute Given Values:**
- Length \( l = 13 \, \text{cm} \)
- Width \( w = 9 \, \text{cm} \)
- Now plug in the values and the rates of change into the differentiated formula.
\[
\frac{dA}{dt} = (13 \, \text{cm}) \times (3 \, \text{cm/s})](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11e77a1c-59f5-4335-a4e9-7f9e80d9c490%2Fb0049ffd-b143-4f42-8890-0d69fbb0ac8d%2Ff0mx59pf.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Calculus Problem: Related Rates
**Problem Statement:**
The length of a rectangle is increasing at a rate of \(6 \, \text{cm/s}\) and its width is increasing at a rate of \(3 \, \text{cm/s}\). When the length is \(13 \, \text{cm}\) and the width is \(9 \, \text{cm}\), how fast is the area of the rectangle increasing?
\[ \text{Rate of increase of area} = \boxed{ \text{cm}^2/\text{s} } \]
---
**Explanation:**
This problem involves related rates, a common type of problem in calculus where we determine how one quantity changes as another quantity changes. To solve this problem, you must understand how the dimensions of a geometric shape (in this case, a rectangle) relate to its area and how changes in these dimensions affect the area over time.
1. **Identify Variables:**
- Let \( l \) represent the length of the rectangle.
- Let \( w \) represent the width of the rectangle.
- Let \( A \) represent the area of the rectangle.
2. **Given Rates:**
- The rate of change of length \( \frac{dl}{dt} = 6 \, \text{cm/s} \)
- The rate of change of width \( \frac{dw}{dt} = 3 \, \text{cm/s} \)
3. **Formula for the Area of Rectangle:**
- The area of the rectangle is \( A = l \times w \)
4. **Differentiate with Respect to Time:**
- Use the product rule to differentiate the area with respect to time.
\[
\frac{dA}{dt} = \frac{d}{dt} (l \times w) = l \frac{dw}{dt} + w \frac{dl}{dt}
\]
5. **Substitute Given Values:**
- Length \( l = 13 \, \text{cm} \)
- Width \( w = 9 \, \text{cm} \)
- Now plug in the values and the rates of change into the differentiated formula.
\[
\frac{dA}{dt} = (13 \, \text{cm}) \times (3 \, \text{cm/s})
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